Simplify the following expression and state the condition under which the simplification is valid. You can assume that $z \neq 0$. $x = \dfrac{z}{z(4z + 7)} \div \dfrac{-9}{3(4z + 7)} $
Dividing by an expression is the same as multiplying by its inverse. $x = \dfrac{z}{z(4z + 7)} \times \dfrac{3(4z + 7)}{-9} $ When multiplying fractions, we multiply the numerators and the denominators. $x = \dfrac{ z \times 3(4z + 7) } { z(4z + 7) \times -9 } $ $ x = \dfrac{3z(4z + 7)}{-9z(4z + 7)} $ We can cancel the $4z + 7$ so long as $4z + 7 \neq 0$ Therefore $z \neq -\dfrac{7}{4}$ $x = \dfrac{3z \cancel{(4z + 7})}{-9z \cancel{(4z + 7)}} = -\dfrac{3z}{9z} = -\dfrac{1}{3} $